Does there exist a quasi-isometric embedding $$MCG(S) \to (\mathrm{Teich}(S), d)$$ for $d$ any "known" distance on the Teichmuller space (i.e. Teichmuller, Weil-Petersson, Thurston...) ?
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$\begingroup$ This is not exactly the answer you want but you can find some related results here <arxiv.org/abs/math/0701719> $\endgroup$– CuspCommented Oct 13, 2013 at 3:52
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1$\begingroup$ I believe the answer is no for Teichmuller and Weil-Petersson: I think this follows from work of Minsky, Rafi, and Brock, who found combinatorial models for measuring distance in the mapping class group, Teichmuller metric, and Weil-Petersson metric respectively. $\endgroup$– Ian AgolCommented Oct 13, 2013 at 4:14
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$\begingroup$ @Ian: I am not sure, since a qi embedding might be unrelated to the standard action of the mcg. $\endgroup$– MishaCommented Oct 13, 2013 at 14:29
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$\begingroup$ Ok, good point Misha! I implicitly assumed he was looking for an equivariant quasi-isometry, so whether the orbit of a point under the mcg is a q.i. embedding. $\endgroup$– Ian AgolCommented Oct 13, 2013 at 16:48
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1 Answer
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A result of Behrstock and Minsky (cf. Hamenstadt too) implies that the rank of mapping class groups is the maximal rank of abelian subgroups, which is $3g+p-3$ for a connected hyperbolic surface of genus $g$ with $p$ boundary components. The rank of Teichmuller space with respect to the Weil-Petersson metric they show is $\lfloor \frac{3g+p-2}{2}\rfloor < 3g-3+p$, so there is no q.i. embedding of the mapping class group to Teichmuller space with the Weil-Petersson metric.
On the other hand, Eskin-Masur-Rafi show that the rank of the Teichmuller metric is equal to that of the mapping class group. They point out that the orbit of a point under the mapping class group is not a quasi-isometric embedding. But this leaves open whether there might be a non-equivariant q.i. embedding.
answered Oct 13, 2013 at 17:15